Consider the electronic configuration of the following elements:A:1s2 2s2 2p6 3s1A : 1s2 2s2 2p6 3s1B:2s2 2s2 2p6 3s2 3p5B : 2s2 2s2 2p6 3s2 3p5C:1s2 2s2 2p6 3s2 3p2C : 1s2 2s2 2p6 3s2 3p2D: 1s2 2s2 2p5D : 1s2 2s2 2p5The element having a maximum difference between the first and second ionization energy is:1.  A2.  C3.  B4.  D

To determine the element with the maximum difference between the first and second ionization energy, we need to compare the electronic configurations of the given elements.

The first ionization energy is the energy required to remove one electron from an atom in its neutral state, while the second ionization energy is the energy required to remove a second electron from the same atom.

Looking at the electronic configurations of the elements:

A: 1s2 2s2 2p6 3s1 B: 2s2 2s2 2p6 3s2 3p5 C: 1s2 2s2 2p6 3s2 3p2 D: 1s2 2s2 2p5

We can see that element B has the highest difference between the first and second ionization energy. This is because it has a completely filled 3p subshell (3p5) in its neutral state, and removing an electron from this subshell requires a significantly higher amount of energy compared to removing an electron from the 3s subshell.

Therefore, the element with the maximum difference between the first and second ionization energy is B.